Question

A block moving horizontally on a smooth surface with a speed of $40\, m / s$ splits into two parts with masses in the ratio of $1: 2$. If the smaller part moves at $60 \,m / s$ in the same direction, then the fractional change in kinetic energy is :-

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{8}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$3 MV _{0}=2 MV _{2}+ MV _{1}$
$3 V _{0}=2 V _{2}+ V _{1}$
$120=2 V _{2}+60 $
$\Rightarrow V _{2}=30 \,m / s$
$\frac{\Delta K . E .}{ K . E .} =\frac{\frac{1}{2} MV _{1}^{2}+\frac{1}{2} 2 MV _{2}^{2}-\frac{1}{2} 3 MV _{0}^{2}}{\frac{1}{2} 3 MV _{0}^{2}} $
$=\frac{ V _{1}^{2}+2 V _{2}^{2}-3 V _{0}^{2}}{3 V _{0}^{2}} $
$=\frac{3600+1800-4800}{4800} $
$=\frac{1}{8} $